# System of two equations with two unknowns and one parameter

I need to solve this system of two equations for $$x\in[0,0.5]$$ and $$y\in[0.5,1]$$, leaving $$b\in[-1,1]$$ as a parameter in the solutions.

$$\begin{cases} 6x^2 b-y(2+(-2+y)b)-2x(-2+(2+y)b)=0\\ -2+y(4-6b)+b-x^2 b+6y^2 b-2x(1+(-1+y)b)=0 \end{cases}$$

WolframAlpha finds only one real solution for $$b=0$$, $$x=1/3$$ and $$y=2/3$$, but there are others.

What I would like to find, if possible, are two functions $$y(b)$$ and $$x(b)$$ that yield the solutions.

See WolframAlpha

Geogebra plot: https://www.geogebra.org/calculator/g3592hwy

The only way I was actually able to solve this system is by setting the parameter $$b$$ to a given number, reducing the system of equation and then solving for $$x$$ and $$y$$. See example: for $$b=1$$ the system of equations becomes

$$\begin{cases} 6x^2-2xy-y^2=0\\ 6y^2-2xy-x^2-2y-1=0 \end{cases}$$

whose solutions are $$x\approx0.457849, y\approx0.753505$$.

Using resultants we see that the general solution with a parameter $$b$$ is given by $$x=\frac{1029 b^2y^3 - 1764b^2y^2 + 938b^2y - 140b^2 + 980by^2 - 1188by + 348b + 212y - 136}{4(7b^2 + 10b + 4)}$$ where $$y$$ is a root of the quartic equation $$1029b^3y^4 - 2352b^3y^3 + 1876b^3y^2 - 608b^3y + 68b^3 + 1568b^2y^3 - 2856b^2y^2 + 1616b^2y - 272b^2 + 732by^2 - 960by + 304b + 96y - 64=0$$ For $$b=0$$ we obtain $$y=2/3$$ and hence $$x=1/3$$.

• I don't think I've understood your solution. Lets say $b=0.5$, how do you find $x$ and $y$? Aug 3, 2020 at 13:22
• Just solve the quartic equation for $y$ with $b=1/2$ and then insert this into the first equation $x=f(y,b)$. I obtain $4$ possible solutions for $x$ (and so for $y$), namely $x=0.394225559926, x= - 0.221896858989, x= - 0.739503037665, x= - 0.766158996606$. Aug 3, 2020 at 13:31
• Ok yes, I got the point. This is not exactly the solution I was hoping for, but it does the trick, thanks Aug 3, 2020 at 13:40
• But now you have the functions $x(b)$ and $y(b)$ "that yield the solutions", as you requested. The only problem is, that we need to solve a quartic equation. There is a general formula, but as Yves said, it is terrible. Aug 3, 2020 at 13:41

You can eliminate $$y^2$$ from the two equations, then express $$y$$ in terms of $$x$$. Plug in one of the initial equations. It seems that you will get a quartic polynomial in $$x$$. There are analytical formulas for the roots of a quartic, but they are a little scary.

• I can eliminate $y^2$? Why? Aug 3, 2020 at 13:20
• @user_231578: this is said in the first sentence.
– user65203
Aug 3, 2020 at 13:23
• I can't understand the reason why $y^2$ can be eliminated Aug 3, 2020 at 13:27
• @user_231578: eliminate it and you'll see. (Or do you ignore the meaning of "eliminate" ?)
– user65203
Aug 3, 2020 at 13:41
• I tried to eliminate the $y^2$'s and then solved on WolframAlpha. The solutions it gives are incorrect. It may be that I don't understand what "eliminate" means. Aug 3, 2020 at 19:33